if-M-is-a-point-on-the-line-y-x-and-points-P-0-1-Q-2-0-are-such-that-PM-PQ-is-minimum-then-find-P-

Question Number 148993 by gsk2684 last updated on 02/Aug/21

Commented by mr W last updated on 02/Aug/21

$${i}\:{think}\:{the}\:{question}\:{should}\:{be} \\ $$$${find}\:{M}\:{such}\:{that}\:{PM}+{QM}\:{is} \\ $$$${minimum}. \\ $$

Commented by gsk2684 last updated on 02/Aug/21

$${yes}\: \\ $$$${typographical}\:{error} \\ $$$${solution}\:{please} \\ $$

Commented by bramlexs22 last updated on 02/Aug/21

if PM+MQ minimum ⇒M? let f(x)=(√(x^2 +(x−1)^2 )) +(√((x−2)^2 +x^2 )) f(x)_(min) if points M,P and Q colinear ⇒so we get m_(PQ) =m_(MQ) ⇒((1−0)/(0−2)) = ((x−0)/(x−2)) ⇒−(1/2)=(x/(x−2)) ⇒−x+2=2x ; x=(2/3) thus M((2/3),(2/3))

$$\mathrm{if}\:\mathrm{PM}+\mathrm{MQ}\:\mathrm{minimum}\:\Rightarrow\mathrm{M}? \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}^{\mathrm{2}} +\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\sqrt{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{f}\left(\mathrm{x}\right)_{\mathrm{min}} \:\mathrm{if}\:\mathrm{points}\:\mathrm{M},\mathrm{P}\:\mathrm{and}\:\mathrm{Q}\:\mathrm{colinear} \\ $$$$\Rightarrow\mathrm{so}\:\mathrm{we}\:\mathrm{get}\:\mathrm{m}_{\mathrm{PQ}} =\mathrm{m}_{\mathrm{MQ}} \\ $$$$\Rightarrow\frac{\mathrm{1}−\mathrm{0}}{\mathrm{0}−\mathrm{2}}\:=\:\frac{\mathrm{x}−\mathrm{0}}{\mathrm{x}−\mathrm{2}} \\ $$$$\Rightarrow−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{x}}{\mathrm{x}−\mathrm{2}}\: \\ $$$$\Rightarrow−\mathrm{x}+\mathrm{2}=\mathrm{2x}\:;\:\mathrm{x}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{thus}\:\mathrm{M}\left(\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$

Commented by bramlexs22 last updated on 02/Aug/21

Commented by gsk2684 last updated on 04/Aug/21

$${i}\:{have}\:{found}\:{that}\:{feet}\:{of}\:{perpendiculars} \\ $$$${from}\:{given}\:{points}\:{on}\:{to}\:{the}\:{given}\: \\ $$$${line}\: \\ $$$${and}\:{using}\:{these}\:{right}\:{angled}\:{triangles} \\ $$$${can}\:{i}\:{go}\:{further}\:{to}\:{find}\:{M}\:{such}\:{that} \\ $$$${PM}+{QM}\:{to}\:{be}\:{minimum} \\ $$$${help}\:{me}\: \\ $$$$ \\ $$

Answered by iloveisrael last updated on 02/Aug/21

M(x,x)⇒PM+PQ=(√(x^2 +(x−1)^2 )) +(√(2^2 +(−1)^2 )) = (√5) +(√(2x^2 −2x+1)) minimum when 4x−2=0 ; x=(1/2) then M((1/2),(1/2)) why P ?

$$\mathrm{M}\left(\mathrm{x},\mathrm{x}\right)\Rightarrow\mathrm{PM}+\mathrm{PQ}=\sqrt{\mathrm{x}^{\mathrm{2}} +\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\sqrt{\mathrm{2}^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$$$=\:\sqrt{\mathrm{5}}\:+\sqrt{\mathrm{2x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{1}}\: \\ $$$$\mathrm{minimum}\:\mathrm{when}\:\mathrm{4x}−\mathrm{2}=\mathrm{0}\:;\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{then}\:\mathrm{M}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{why}\:\mathrm{P}\:? \\ $$

Commented by gsk2684 last updated on 02/Aug/21

$${thank}\:{you}\: \\ $$

Answered by Kamel last updated on 02/Aug/21

if M is a point on the line y=x and points P(0,1),Q(2,0) are such that PM+PQ is minimum then find P PM+QM minimum⇒M∈(PQ) /M(x,x) (PQ): y=ax+1⇒(PQ):y=−(1/2)x+1 M∈(PQ)⇔x=y=−(1/2)x+1⇒x=y=(2/3)

$${if}\:{M}\:{is}\:{a}\:{point}\:{on}\:{the}\:{line}\:{y}={x}\:{and} \\ $$$${points}\:{P}\left(\mathrm{0},\mathrm{1}\right),{Q}\left(\mathrm{2},\mathrm{0}\right)\:{are}\:{such}\:{that} \\ $$$${PM}+{PQ}\:{is}\:{minimum}\:{then}\:{find}\:{P} \\ $$$${PM}+{QM}\:{minimum}\Rightarrow{M}\in\left({PQ}\right)\:/{M}\left({x},{x}\right) \\ $$$$\left({PQ}\right):\:{y}={ax}+\mathrm{1}\Rightarrow\left({PQ}\right):{y}=−\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{1} \\ $$$${M}\in\left({PQ}\right)\Leftrightarrow{x}={y}=−\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{1}\Rightarrow{x}={y}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$

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